@heap-data-structure/binomial-heap/dist/index.cjs.map

Binomial heap data structures for JavaScript

{"version":3,"file":"index.cjs","sources":["../src/BinomialTree.js","../src/BinomialTreeWithParent.js","../src/LazyNode.js","../src/LazyStack.js","../src/BinomialHeap.js","../src/LazyBinomialHeap.js"],"sourcesContent":["export default function BinomialTree(value, children) {\n\tthis.value = value;\n\tthis.children = children;\n}\n\n/**\n * /!\\ Can only be used to merge two trees of the same rank.\n * /!\\ Modifies both trees\n */\n\nBinomialTree.prototype.merge = function (predicate, other) {\n\tif (predicate(this.value, other.value) <= 0) {\n\t\tthis.children = this.children.concat(other);\n\t\treturn this;\n\t}\n\n\tother.children = other.children.concat(this);\n\treturn other;\n};\n\n/**\n * Method used to reset a tree element in order to reuse it\n * somewhere else, e.g. insert it back in the same or a new\n * heap.\n */\n\nBinomialTree.prototype.detach = function () {\n\tthis.children.splice(0);\n\treturn this;\n};\n\nBinomialTree.prototype.setparent = function (_parent) {\n\t// We do not keep track of any parent here\n};\n","export default function BinomialTreeWithParent(value, children) {\n\tthis.value = value;\n\tthis.children = children;\n\tthis.parent = null;\n}\n\nBinomialTreeWithParent.prototype.rank = function () {\n\treturn this.children.length;\n};\n\n/**\n * /!\\ Can only be used to merge two trees of the same rank.\n * /!\\ Modifies both trees\n */\n\nBinomialTreeWithParent.prototype.merge = function (predicate, other) {\n\tif (predicate(this.value, other.value) <= 0) {\n\t\tthis.children = this.children.concat(other);\n\t\tother.parent = this;\n\t\treturn this;\n\t}\n\n\tother.children = other.children.concat(this);\n\tthis.parent = other;\n\treturn other;\n};\n\n/**\n * Method used to reset a tree element in order to reuse it\n * somewhere else, e.g. insert it back in the same or a new\n * heap.\n */\n\nBinomialTreeWithParent.prototype.detach = function () {\n\tthis.children.splice(0);\n\tthis.parent = null;\n\treturn this;\n};\n\nBinomialTreeWithParent.prototype.setparent = function (parent) {\n\tthis.parent = parent;\n};\n","export default function LazyNode(value, next) {\n\tthis.value = value;\n\tthis.next = next;\n}\n","import LazyNode from './LazyNode.js';\n\n/**\n * LazyStack#peek only valid if LazyStack#empty is false.\n * LazyStack#shift only valid if LazyStack#empty is false.\n */\n\nexport default function LazyStack() {\n\tthis.top = null;\n\tthis.bottom = null;\n}\n\nLazyStack.prototype.empty = function () {\n\treturn this.top === null;\n};\n\nLazyStack.prototype.push = function (value) {\n\tthis.top = new LazyNode(value, this.top);\n\n\tif (this.bottom === null) this.bottom = this.top;\n};\n\n/**\n * Only valid if LazyStack#empty is false.\n */\n\nLazyStack.prototype.pop = function () {\n\tconst value = this.top.value;\n\n\tthis.top = this.top.next;\n\n\tif (this.top === null) this.bottom = null;\n\n\treturn value;\n};\n\nLazyStack.prototype.meld = function (other) {\n\tif (this.bottom === null) this.top = other.top;\n\telse this.bottom.next = other.top;\n\n\tthis.bottom = other.bottom;\n};\n","export default function BinomialHeap(BinomialTree) {\n\tconst binomial_heap_push = function (compare, list, tree, rank) {\n\t\t// Ensures list has at least rank cells\n\n\t\tlet i = rank - list.length;\n\n\t\twhile (i-- > 0) {\n\t\t\tlist.push(null);\n\t\t}\n\n\t\t// Loop invariant\n\t\t// tree and list[i] have the same rank\n\n\t\tconst len = list.length;\n\n\t\tfor (i = rank; i < len && list[i] !== null; ++i) {\n\t\t\t// There is already a tree with this rank\n\n\t\t\ttree = tree.merge(compare, list[i]);\n\t\t\tlist[i] = null;\n\t\t}\n\n\t\t// Do not forget to append null if\n\t\t// we are lacking space\n\n\t\tif (i === len) {\n\t\t\tlist.push(null);\n\t\t}\n\n\t\t// Cell is empty\n\t\t// we can just put the new tree here\n\n\t\tlist[i] = tree;\n\t};\n\n\tconst merge = function (compare, list, other) {\n\t\tif (other.length === 0) {\n\t\t\treturn;\n\t\t}\n\n\t\t// Merging two binomial heaps is like\n\t\t// adding two little endian integers\n\t\t// so, we first make sure that we have\n\t\t// enough place to store the result\n\n\t\tlet i = other.length - list.length;\n\n\t\twhile (i-- > 0) {\n\t\t\tlist.push(null);\n\t\t}\n\n\t\tlet carry = null;\n\n\t\tconst len = list.length;\n\n\t\t// Remember len >= other.length\n\n\t\tfor (i = 0; i < len; ++i) {\n\t\t\t// Other[i] can be either null or not\n\t\t\t// list[i] can be either null or not\n\t\t\t// carry can be either null or not\n\t\t\t// --> 2^3 = 8 possibilities\n\t\t\t//\n\t\t\t//    null ? | other[i] | list[i] | carry\n\t\t\t// ---------------------------------------\n\t\t\t//     (0)   |    no    |     no  |   no\n\t\t\t//     (1)   |    no    |     no  |  yes\n\t\t\t//     (2)   |    no    |    yes  |   no\n\t\t\t//     (3)   |    no    |    yes  |  yes\n\t\t\t//     (4)   |   yes    |     no  |   no\n\t\t\t//     (5)   |   yes    |     no  |  yes\n\t\t\t//     (6)   |   yes    |    yes  |   no\n\t\t\t//     (7)   |   yes    |    yes  |  yes\n\n\t\t\tif (i >= other.length || other[i] === null) {\n\t\t\t\tif (carry !== null) {\n\t\t\t\t\t// (6) other[i] = null and list[i] = null and carry != null\n\t\t\t\t\t// --> put carry in current cell\n\n\t\t\t\t\tif (list[i] === null) {\n\t\t\t\t\t\tlist[i] = carry;\n\t\t\t\t\t\tcarry = null;\n\t\t\t\t\t}\n\n\t\t\t\t\t// (4) other[i] = null and list[i] != null and carry != null\n\t\t\t\t\t// --> merge carry with current cell\n\t\t\t\t\telse {\n\t\t\t\t\t\tcarry = carry.merge(compare, list[i]);\n\t\t\t\t\t\tlist[i] = null;\n\t\t\t\t\t}\n\t\t\t\t}\n\n\t\t\t\t// We do not need to do anything for\n\t\t\t\t// those 2 cases (carry and other[i] are null).\n\t\t\t\t// ==\n\t\t\t\t// (5) other[i] = null and list[i] != null and carry = null\n\t\t\t\t// (7) other[i] = null and list[i] = null and carry = null\n\t\t\t}\n\n\t\t\t// (0) other[i] != null and list[i] != null and carry != null\n\t\t\t// (2) other[i] != null and list[i] = null and carry != null\n\t\t\t// --> merge carry with other[i]\n\t\t\telse if (carry !== null) {\n\t\t\t\tcarry = carry.merge(compare, other[i]);\n\t\t\t}\n\n\t\t\t// (1) other[i] != null and list[i] != null and carry = null\n\t\t\t// --> merge current cell with other[i]\n\t\t\telse if (list[i] !== null) {\n\t\t\t\tcarry = list[i].merge(compare, other[i]);\n\t\t\t\tlist[i] = null;\n\t\t\t}\n\n\t\t\t// (3) other[i] != null and list[i] = null and carry = null\n\t\t\t// --> put other[i] in list\n\t\t\telse {\n\t\t\t\tlist[i] = other[i];\n\t\t\t}\n\t\t}\n\n\t\t// Do not forget to append last carry\n\n\t\tif (carry !== null) {\n\t\t\tlist.push(carry);\n\t\t}\n\t};\n\n\tconst find_min_index = function (compare, list, j, len) {\n\t\t// There MUST be at least one\n\t\t// non null element in this list\n\t\t// we look for the first one\n\n\t\tfor (; j < len - 1 && list[j] === null; ++j);\n\n\t\t// Here j is necessarily < len\n\t\t// and list[j] is non null\n\n\t\tlet i = j;\n\t\tlet opt = list[j].value;\n\n\t\t// We lookup remaining elements to see if there\n\t\t// is not a better candidate\n\n\t\tfor (++j; j < len; ++j) {\n\t\t\tconst item = list[j];\n\n\t\t\tif (item !== null) {\n\t\t\t\tconst candidate = item.value;\n\n\t\t\t\tif (compare(candidate, opt) < 0) {\n\t\t\t\t\ti = j;\n\t\t\t\t\topt = candidate;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\n\t\treturn i;\n\t};\n\n\tconst remove_head_at_index = function (compare, list, i, len) {\n\t\tconst orphans = list[i].children;\n\t\tlist[i] = null;\n\n\t\tchange_parent(null, orphans);\n\n\t\t// We just removed the ith element\n\t\t// if list[i] is the last cell\n\t\t// of list we can drop it\n\n\t\tif (i === len - 1) {\n\t\t\tlist.pop();\n\t\t}\n\n\t\t// We merge back the children of\n\t\t// the removed tree into the heap\n\n\t\tmerge(compare, list, orphans);\n\t};\n\n\tconst binomial_heap_pop = function (compare, list) {\n\t\tconst len = list.length;\n\n\t\tconst i = find_min_index(compare, list, 0, len);\n\n\t\tconst tree = list[i];\n\n\t\tremove_head_at_index(compare, list, i, len);\n\n\t\treturn tree;\n\t};\n\n\tconst change_parent = function (parent, children) {\n\t\tconst len = children.length;\n\n\t\tfor (let i = 0; i < len; ++i) {\n\t\t\tchildren[i].setparent(parent);\n\t\t}\n\t};\n\n\tconst shift_up = function (tree, parent) {\n\t\t// Console.log( \"tree\", tree.value );\n\t\t// console.log( \"parent\", parent.value );\n\n\t\t// Here, we cannot just swap values as it would invalidate\n\t\t// externally stored references.\n\t\t// Instead, we swap children lists and update references\n\t\t// between the tree and its parent.\n\t\t// Then we update and return the new tree's parent.\n\n\t\t// console.log( \"tree.children\", tree.children );\n\t\t// console.log( \"parent.children\", parent.children );\n\n\t\tconst tmp = parent.children;\n\t\tparent.children = tree.children;\n\t\ttree.children = tmp;\n\n\t\tconst i = parent.rank();\n\n\t\t// Console.log( tree.children, i );\n\n\t\ttree.children[i] = parent;\n\n\t\ttree.parent = parent.parent;\n\n\t\tchange_parent(tree, tree.children);\n\t\tchange_parent(parent, parent.children);\n\n\t\t// Console.log( \"tree.children\", tree.children );\n\t\t// console.log( \"parent.children\", parent.children );\n\n\t\treturn tree.parent;\n\t};\n\n\tconst percolate_up = function (list, tree) {\n\t\tlet parent = tree.parent;\n\n\t\tif (parent !== null) {\n\t\t\twhile (true) {\n\t\t\t\tparent = shift_up(tree, parent);\n\n\t\t\t\tif (parent === null) {\n\t\t\t\t\tbreak;\n\t\t\t\t}\n\n\t\t\t\t// TODO this call might not be necessary\n\t\t\t\tparent.children[tree.rank()] = tree;\n\t\t\t}\n\n\t\t\tlist[tree.rank()] = tree;\n\t\t}\n\t};\n\n\tconst decreasekey = function (compare, list, tree, value) {\n\t\ttree.value = value;\n\t\tlet parent = tree.parent;\n\n\t\tif (parent !== null) {\n\t\t\twhile (true) {\n\t\t\t\tconst d = compare(value, parent.value);\n\n\t\t\t\tif (d >= 0) {\n\t\t\t\t\treturn;\n\t\t\t\t}\n\n\t\t\t\tparent = shift_up(tree, parent);\n\n\t\t\t\tif (parent === null) {\n\t\t\t\t\tbreak;\n\t\t\t\t}\n\n\t\t\t\t// TODO this call should be in if ( d >= 0 )\n\t\t\t\tparent.children[tree.rank()] = tree;\n\t\t\t}\n\n\t\t\tlist[tree.rank()] = tree;\n\t\t}\n\t};\n\n\tconst deletetree = function (compare, list, tree) {\n\t\tpercolate_up(list, tree);\n\n\t\tremove_head_at_index(compare, list, tree.rank(), list.length);\n\n\t\ttree.detach();\n\t};\n\n\tconst Heap = function (compare) {\n\t\t// The compare function to use to compare values\n\n\t\tthis.compare = compare;\n\n\t\t// Number of elements in this heap\n\n\t\tthis.length = 0;\n\n\t\t// List of binomial trees\n\n\t\tthis.list = [];\n\t};\n\n\tHeap.prototype.head = function () {\n\t\tif (this.length === 0) {\n\t\t\treturn undefined;\n\t\t}\n\n\t\tconst i = find_min_index(this.compare, this.list, 0, this.list.length);\n\n\t\tconst tree = this.list[i];\n\n\t\treturn tree.value;\n\t};\n\n\tHeap.prototype.headreference = function () {\n\t\tif (this.length === 0) {\n\t\t\treturn null;\n\t\t}\n\n\t\tconst i = find_min_index(this.compare, this.list, 0, this.list.length);\n\n\t\tconst tree = this.list[i];\n\n\t\treturn tree;\n\t};\n\n\tHeap.prototype.pop = function () {\n\t\tif (this.length === 0) {\n\t\t\treturn undefined;\n\t\t}\n\n\t\t--this.length;\n\n\t\treturn binomial_heap_pop(this.compare, this.list).value;\n\t};\n\n\tHeap.prototype.popreference = function () {\n\t\tif (this.length === 0) {\n\t\t\treturn null;\n\t\t}\n\n\t\t--this.length;\n\n\t\treturn binomial_heap_pop(this.compare, this.list).detach();\n\t};\n\n\tHeap.prototype.push = function (value) {\n\t\t// Push a new tree of rank 0\n\n\t\tconst tree = new BinomialTree(value, []);\n\n\t\tthis.pushreference(tree);\n\n\t\treturn tree;\n\t};\n\n\tHeap.prototype.pushreference = function (tree) {\n\t\t++this.length;\n\n\t\t// Push an existing tree of rank 0\n\n\t\tbinomial_heap_push(this.compare, this.list, tree, 0);\n\t};\n\n\tHeap.prototype.merge = function (other) {\n\t\tmerge(this.compare, this.list, other.list);\n\n\t\tthis.length += other.length;\n\n\t\treturn this;\n\t};\n\n\tHeap.prototype.update = function (tree, value) {\n\t\tconst d = this.compare(value, tree.value);\n\n\t\tif (d < 0) {\n\t\t\tthis.decreasekey(tree, value);\n\t\t} else if (d > 0) {\n\t\t\tthis.increasekey(tree, value);\n\t\t} else {\n\t\t\t// D === 0 does not imply tree.value === value\n\n\t\t\ttree.value = value;\n\t\t}\n\t};\n\n\tHeap.prototype.decreasekey = function (tree, value) {\n\t\tdecreasekey(this.compare, this.list, tree, value);\n\t};\n\n\tHeap.prototype.increasekey = function (tree, value) {\n\t\tdeletetree(this.compare, this.list, tree);\n\n\t\ttree.value = value;\n\n\t\tbinomial_heap_push(this.compare, this.list, tree, 0);\n\t};\n\n\tHeap.prototype.delete = function (tree) {\n\t\t--this.length;\n\n\t\tdeletetree(this.compare, this.list, tree);\n\t};\n\n\treturn Heap;\n}\n","import LazyStack from './LazyStack.js';\n\nexport default function LazyBinomialHeap(BinomialTree) {\n\tconst lazy_binomial_heap_push = function (lazy, tree, rank) {\n\t\t// Lightweight binomial heap containing a unique tree\n\t\tconst sequence = [];\n\n\t\t// Offset tree by its rank\n\t\tlet i = rank;\n\n\t\twhile (i--) {\n\t\t\tsequence.push(null);\n\t\t}\n\n\t\tsequence.push(tree);\n\n\t\t// Do not merge the generated sequence immediately\n\t\tlazy.push(sequence);\n\t};\n\n\tconst merge = function (compare, list, other) {\n\t\tif (other.length === 0) {\n\t\t\treturn;\n\t\t}\n\n\t\t// Merging two binomial heaps is like\n\t\t// adding two little endian integers\n\t\t// so, we first make sure that we have\n\t\t// enough place to store the result\n\n\t\tlet i = other.length - list.length;\n\n\t\twhile (i-- > 0) {\n\t\t\tlist.push(null);\n\t\t}\n\n\t\tlet carry = null;\n\n\t\tconst len = list.length;\n\n\t\t// Remember len >= other.length\n\n\t\tfor (i = 0; i < len; ++i) {\n\t\t\t// Other[i] can be either null or not\n\t\t\t// list[i] can be either null or not\n\t\t\t// carry can be either null or not\n\t\t\t// --> 2^3 = 8 possibilities\n\t\t\t//\n\t\t\t//    null ? | other[i] | list[i] | carry\n\t\t\t// ---------------------------------------\n\t\t\t//     (0)   |    no    |     no  |   no\n\t\t\t//     (1)   |    no    |     no  |  yes\n\t\t\t//     (2)   |    no    |    yes  |   no\n\t\t\t//     (3)   |    no    |    yes  |  yes\n\t\t\t//     (4)   |   yes    |     no  |   no\n\t\t\t//     (5)   |   yes    |     no  |  yes\n\t\t\t//     (6)   |   yes    |    yes  |   no\n\t\t\t//     (7)   |   yes    |    yes  |  yes\n\n\t\t\tif (i >= other.length || other[i] === null) {\n\t\t\t\tif (carry !== null) {\n\t\t\t\t\t// (6) other[i] = null and list[i] = null and carry != null\n\t\t\t\t\t// --> put carry in current cell\n\n\t\t\t\t\tif (list[i] === null) {\n\t\t\t\t\t\tlist[i] = carry;\n\t\t\t\t\t\tcarry = null;\n\t\t\t\t\t}\n\n\t\t\t\t\t// (4) other[i] = null and list[i] != null and carry != null\n\t\t\t\t\t// --> merge carry with current cell\n\t\t\t\t\telse {\n\t\t\t\t\t\tcarry = carry.merge(compare, list[i]);\n\t\t\t\t\t\tlist[i] = null;\n\t\t\t\t\t}\n\t\t\t\t}\n\n\t\t\t\t// We do not need to do anything for\n\t\t\t\t// those 2 cases (carry and other[i] are null).\n\t\t\t\t// ==\n\t\t\t\t// (5) other[i] = null and list[i] != null and carry = null\n\t\t\t\t// (7) other[i] = null and list[i] = null and carry = null\n\t\t\t}\n\n\t\t\t// (0) other[i] != null and list[i] != null and carry != null\n\t\t\t// (2) other[i] != null and list[i] = null and carry != null\n\t\t\t// --> merge carry with other[i]\n\t\t\telse if (carry !== null) {\n\t\t\t\tcarry = carry.merge(compare, other[i]);\n\t\t\t}\n\n\t\t\t// (1) other[i] != null and list[i] != null and carry = null\n\t\t\t// --> merge current cell with other[i]\n\t\t\telse if (list[i] !== null) {\n\t\t\t\tcarry = list[i].merge(compare, other[i]);\n\t\t\t\tlist[i] = null;\n\t\t\t}\n\n\t\t\t// (3) other[i] != null and list[i] = null and carry = null\n\t\t\t// --> put other[i] in list\n\t\t\telse {\n\t\t\t\tlist[i] = other[i];\n\t\t\t}\n\t\t}\n\n\t\t// Do not forget to append last carry\n\n\t\tif (carry !== null) {\n\t\t\tlist.push(carry);\n\t\t}\n\t};\n\n\tconst lazy_binomial_heap_pop = function (compare, list, lazy) {\n\t\t// Amortized merge of stored values\n\n\t\twhile (!lazy.empty()) merge(compare, list, lazy.pop());\n\n\t\t// Standard O(log n) optimum search method\n\n\t\tconst len = list.length;\n\n\t\t// There MUST be at least one\n\t\t// non null element in this list\n\t\t// we look for the first one\n\n\t\tlet j = 0;\n\t\tfor (; j < len - 1 && list[j] === null; ++j);\n\n\t\t// Here j is necessarily < len\n\t\t// and list[j] is non null\n\n\t\tlet i = j;\n\t\tlet opt = list[j].value;\n\n\t\t// We lookup remaining elements to see if there\n\t\t// is not a better candidate\n\n\t\tfor (++j; j < len; ++j) {\n\t\t\tconst item = list[j];\n\n\t\t\tif (item !== null) {\n\t\t\t\tconst candidate = item.value;\n\n\t\t\t\tif (compare(candidate, opt) < 0) {\n\t\t\t\t\ti = j;\n\t\t\t\t\topt = candidate;\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\n\t\tconst orphan = list[i].children;\n\t\tlist[i] = null;\n\n\t\t// We just removed the ith element\n\t\t// if list[i] is the last cell\n\t\t// of list we can drop it\n\n\t\tif (i === len - 1) {\n\t\t\tlist.pop();\n\t\t}\n\n\t\t// We store the children in the\n\t\t// lazy list\n\n\t\tlazy.push(orphan);\n\n\t\treturn opt;\n\t};\n\n\tconst Heap = function (compare) {\n\t\t// The compare function to use to compare values\n\n\t\tthis.compare = compare;\n\n\t\t// Number of elements in this heap\n\n\t\tthis.length = 0;\n\n\t\t// List of binomial trees\n\n\t\tthis.list = [];\n\n\t\t// List of binomial heaps waiting to be merged\n\n\t\tthis.lazy = new LazyStack();\n\t};\n\n\tHeap.prototype.pop = function () {\n\t\tif (this.length === 0) {\n\t\t\treturn undefined;\n\t\t}\n\n\t\t--this.length;\n\n\t\treturn lazy_binomial_heap_pop(this.compare, this.list, this.lazy);\n\t};\n\n\tHeap.prototype.push = function (value) {\n\t\t++this.length;\n\n\t\t// Push a new tree of rank 0\n\n\t\treturn lazy_binomial_heap_push(this.lazy, new BinomialTree(value, []), 0);\n\t};\n\n\tHeap.prototype.merge = function (other) {\n\t\tthis.lazy.meld(other.lazy);\n\n\t\tthis.length += other.length;\n\n\t\treturn this;\n\t};\n\n\treturn 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